 
\chapter{Appendix: Integral tables}



%United States Naval Academy Mathematics Department Math Tables, Version 1.0
%1992, Chris Sagovac

%United States Naval Academy Mathematics Department Math Tables, Version 1.1
%2005, David Joyner

%United States Naval Academy Mathematics Department Math Tables, Version 1.2
%2005, David Joyner and Tony Gaglione

%initiate automatic table trigonometry entry numbering
\newcount\trigentryno
\def\trigentrytitle{\noindent\llap{\largestrut T\the\trigentryno.\qquad}}
\def\newtrigentry{\global\advance\trigentryno by 1\trigentrytitle}

%initiate automatic table power series numbering
\newcount\powerentryno
\def\powerentrytitle{\noindent\llap{\largestrut P\the\powerentryno.\qquad}}
\def\newpowerentry{\global\advance\powerentryno by 1\powerentrytitle}

%initiate automatic table integral numbering
\newcount\intentryno
\def\intentrytitle{\noindent\llap{\largestrut I\the\intentryno.\qquad}}
\def\newintentry{\global\advance\intentryno by 1\intentrytitle}

% 14 equations per page?

% \pageno=1
\hsize=4.5truein
\hoffset=0truein
\vsize=7.25truein

\def\dss {\displaystyle}

%\font\lapfont=amsy10     % for old TeX fonts
\font\lapfont=cmsy10
\def\lt{\hbox{\lapfont\char76}}


\newbox\medstrutbox       % strut for first box on page
\setbox\medstrutbox=\hbox{\vrule height14.5pt depth9.5pt width0pt}
\def\medstrut{\relax\ifmmode\copy\medstrutbox\else\unhcopy\medstrutbox\fi}

\newbox\largestrutbox     % strut for entry boxes
\setbox\largestrutbox=\hbox{\vrule height19.5pt depth13.5pt width0pt}
\def\largestrut{\relax\ifmmode\copy\largestrutbox\else\unhcopy\largestrutbox\fi}


\newcount\leqnumno        % equation numbering
\def\leqnumtitle{\noindent{\largestrut\rm \thinspace L\the\leqnumno.}\thinspace} 
\def\leqnum{\global\advance\leqnumno by 1 \leqnumtitle}

\def\frac#1/#2{\leavevmode\kern.1em
\raise.5ex\hbox{\the\scriptfont0 #1}\kern-.1em
/\kern-.15em\lower.25ex\hbox{\the\scriptfont0 #2}}

 
%==========ARC TRIG FUNCTIONS
\def\asin{\mathop{\rm arcsin}\nolimits}
\def\acos{\mathop{\rm arccos}\nolimits}
\def\atan{\mathop{\rm arctan}\nolimits}
\def\asec{\mathop{\rm arcsec}\nolimits}
\def\acsc{\mathop{\rm arccsc}\nolimits}
\def\acot{\mathop{\rm arccot}\nolimits}

%==========INV TRIG FUNCTIONS
\def\isin{\mathop{\rm sin^{-1}}\nolimits}
\def\icos{\mathop{\rm cos^{-1}}\nolimits}
\def\itan{\mathop{\rm tan^{-1}}\nolimits}
\def\isec{\mathop{\rm sec^{-1}}\nolimits}
\def\icsc{\mathop{\rm csc^{-1}}\nolimits}
\def\icot{\mathop{\rm cot^{-1}}\nolimits}

%==========HYPERBOLIC FUNCTIONS
\def\sinh{\mathop{\rm sinh}\nolimits}
\def\cosh{\mathop{\rm cosh}\nolimits}
\def\tanh{\mathop{\rm tanh}\nolimits}
\def\sech{\mathop{\rm sech}\nolimits}
\def\csch{\mathop{\rm csch}\nolimits}
\def\coth{\mathop{\rm coth}\nolimits}

%==========ARC HYPERBOLIC FUNCTIONS
\def\asinh{\mathop{\rm arcsinh}\nolimits}
\def\acosh{\mathop{\rm arccosh}\nolimits}
\def\atanh{\mathop{\rm arctanh}\nolimits}
\def\asech{\mathop{\rm arcsech}\nolimits}
\def\acsch{\mathop{\rm arccsch}\nolimits}
\def\acoth{\mathop{\rm arccoth}\nolimits}

%==========INV HYPERBOLIC FUNCTIONS
\def\isinh{\mathop{\rm sinh^{-1}}\nolimits}
\def\icosh{\mathop{\rm cosh^{-1}}\nolimits}
\def\itanh{\mathop{\rm tanh^{-1}}\nolimits}
\def\isech{\mathop{\rm sech^{-1}}\nolimits}
\def\icsch{\mathop{\rm csch^{-1}}\nolimits}
\def\icoth{\mathop{\rm coth^{-1}}\nolimits}

%font definitions
%
%Laplace Transform L
\font\lapfont=cmsy10
\def\laplace{\hbox{\lapfont\char 76}}
%
%Other fonts
\font\bigr=cmr17
\font\smallr=cmr8
%


\centerline{{\bigr Trigonometry}}
\vskip 24pt
\centerline{{\bf Trigonometric Functions}}
\vskip 18pt
\newtrigentry {\hfill $\dss \sin^2x+\cos^2x=1$\hfill}

\newtrigentry {\hfill $\dss \tan^2x+1=\sec^2x$\hfill}

\newtrigentry {\hfill $\dss \cot^2x+1=\csc^2x$\hfill}

\newtrigentry {\hfill $\dss \sin (x\pm y)=\sin x\,\cos y\pm\cos x\,\sin y$\hfill}

\newtrigentry {\hfill $\dss \cos (x\pm y)=\cos x\,\cos y\mp\sin x\,\sin y$\hfill}

\newtrigentry {\hfill $\dss \tan (x\pm y)={\tan x\pm\tan y\over 1\mp\tan x\,\tan y}$\hfill}

\newtrigentry {\hfill $\dss \tan ({x\over 2})={\sin x\over 1+\cos x}$\hfill}

\newtrigentry {\hfill $\dss \sin (2x) = 2\sin x\,\cos x$\hfill}

\newtrigentry {\hfill $\dss \cos (2x) = \cos^2x-\sin^2x$\hfill}

\newtrigentry {\hfill $\dss \sin^2x=\frac 1/2 (1-\cos (2x))$\hfill}

\newtrigentry {\hfill $\dss \cos^2x=\frac 1/2 (1+\cos (2x))$\hfill}

\newtrigentry {\hfill $\dss \sin x\,\sin y=\frac 1/2 \bigl (\cos (x-y)-\cos (x+y)\bigr )$\hfill}

\newtrigentry {\hfill $\dss \cos x\,\cos y=\frac 1/2 \bigl (\cos (x-y)+\cos (x+y)\bigr )$\hfill}

\newtrigentry {\hfill $\dss \sin x\,\cos y=\frac 1/2 \bigl (\sin (x-y)+\sin (x+y)\bigr )$\hfill}

\newtrigentry {\hfill $\dss c_{1}\cos (\omega t)+c_{2}\sin (\omega t)=A\,\sin (\omega t+\phi),$\hfill}
$$ {\rm where}\,\, A=\sqrt{c_1^2+c_2^2},\ \ \phi=2\atan {c_1\over c_2 +A}$$
\vskip 24pt
\centerline{{\bf Hyperbolic Functions}}
\vskip 18pt
\newtrigentry {\hfill $\dss \cosh x={e^x+e^{-x}\over 2}$\hfill}

\newtrigentry {\hfill $\dss \sinh x={e^x-e^{-x}\over 2}$\hfill}

\newtrigentry {\hfill $\dss \cosh^2 x-\sinh^2 x=1$\hfill}

\newtrigentry {\hfill $\dss \tanh^2 x+\sech^2 x=1$\hfill}

\newtrigentry {\hfill $\dss \coth^2 x-\csch^2 x=1$\hfill}

\newtrigentry {\hfill $\dss \sinh (x\pm y)=\sinh x\,\cosh y\pm\cosh x\,\sinh y$\hfill}

\newtrigentry {\hfill $\dss \cosh (x\pm y)=\cosh x\,\cosh y\pm\sinh x\,\sinh y$\hfill}

\newtrigentry {\hfill $\dss \tanh (x\pm y)={\tanh x\pm\tanh y\over 1\pm\tanh x\,\tanh y}$\hfill}

\newtrigentry {\hfill $\dss \sinh (2x)=2\sinh x\,\cosh x$\hfill}

\newtrigentry {\hfill $\dss \cosh (2x)=\cosh^2 x +\sinh^2 x$\hfill}

\newtrigentry {\hfill $\dss \sinh x\,\sinh y=\frac 1/2\bigl (\cosh (x+y)-\cosh (x-y)\bigr )$\hfill}

\newtrigentry {\hfill $\dss \cosh x\,\cosh y=\frac 1/2\bigl (\cosh (x+y)+\cosh (x-y)\bigr )$\hfill}

\newtrigentry {\hfill $\dss \sinh x\,\cosh y=\frac 1/2\bigl (\sinh (x+y)+\sinh (x-y)\bigr )$\hfill}

\vskip 24pt
\centerline{{\bigr Power Series}}
\vskip 24pt
\newpowerentry {\hfill $\dss e^x=\sum_{n=0}^{\infty}{x^n\over n!}=1+x+{x^2\over 2!}+{x^3\over 3!}+\cdots,\quad -\infty<x<\infty$\hfill}

\newpowerentry {\hfill $\dss \sin x=\sum_{n=0}^{\infty}(-1)^n{x^{2n+1}\over (2n+1)!}=
x-{x^3\over 3!}+{x^5\over 5!}-\cdots,\quad -\infty<x<\infty$\hfill}

\newpowerentry {\hfill $\dss \cos x=\sum_{n=0}^{\infty}(-1)^n{x^{2n}\over (2n)!}=
1-{x^2\over 2!}+{x^4\over 4!}-\cdots,\quad -\infty<x<\infty$\hfill}

\newpowerentry {\hfill $\dss \tan x=x+{x^3\over 3}+{2\over 15}x^5+{17\over 315}x^7+\cdots,\quad -{\pi\over 2}<x<{\pi\over 2}$\hfill}

\newpowerentry {\hfill $\dss {1\over 1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots,\quad
-1<x<1$\hfill}

\newpowerentry {\hfill $\dss \sinh x=\sum_{n=0}^{\infty}{x^{2n+1}\over (2n+1)!}=
x+{x^3\over 3!}+{x^5\over 5!}+\cdots,\quad -\infty<x<\infty$\hfill}

\newpowerentry {\hfill $\dss \cosh x=\sum_{n=0}^{\infty}{x^{2n}\over (2n)!}=
1+{x^2\over 2!}+{x^4\over 4!}+\cdots,\quad -\infty<x<\infty$\hfill}

\newpowerentry {\hfill $\dss \tanh x=x-{x^3\over 3}+{2\over 15}x^5-{17\over 315}x^7+\cdots,\quad -{\pi\over 2}<x<{\pi\over 2}$\hfill}

%\newpowerentry {\hfill $\dss \ln (1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}{x^n\over n}=
%x-{x^2\over 2}+{x^3\over 3}-\cdots,\quad -1<x\le 1$\hfill}

\newpowerentry {\hfill $\dss {f(x) =f(a)+f' (a)(x-a)+{f'' (a)\over 2!}(x-a)^2+
{f''' (a)\over 3!}(x-a)^3+\cdots}$\hfill}

\newpowerentry {Taylor Series with remainder:
\begin{eqnarray*}
f(x) &=\sum_{n=0}^{N}{f^{(n)}(a)\over n!}\,(x-a)^n+R_{N+1}(x),\quad \hbox{\rm where}\\
R_{N+1}(x) &= {f^{(N+1)}(\xi )\over (N+1)!}\,(x-a)^{N+1}\qquad\hbox{\rm for some}\,\,\xi\,\,\hbox{\rm between}\,\, a\,\,\hbox{\rm and}\,\, 
x.
\end{eqnarray*}
}

\vskip 24pt
\centerline {{\bigr Table of Integrals}}
\vskip 24pt
A constant of integration should be added to each formula. The letters $a$,
$b$, $m$, and $n$ denote constants; $u$ and $v$ denote functions of an 
independent variable such as $x$.
\vskip 24pt
\centerline {{\bf Standard Integrals}}
\vskip 18pt
\newintentry {\hfill $\dss \int u^n\, du = {u^{n+1}\over n+1},\quad n\ne -1$\hfill}

\newintentry {\hfill $\dss \int {du\over u}=\ln |u|$\hfill}

\newintentry {\hfill $\dss \int e^u\, du= e^u$\hfill}

\newintentry {\hfill $\dss \int a^u\, du={a^u\over \ln a},\quad a>0$\hfill}

\newintentry {\hfill $\dss \int\cos u\, du=\sin u$\hfill}

\newintentry {\hfill $\dss \int\sin u\, du= -\cos u$\hfill}

\newintentry {\hfill $\dss \int\sec^2u\, du=\tan u$\hfill}

\newintentry {\hfill $\dss \int\csc^2u\, du= -\cot u$\hfill}

\newintentry {\hfill $\dss \int\sec u\tan u\, du=\sec u$\hfill}

\newintentry {\hfill $\dss \int\csc u\cot u\, du = -\csc u$\hfill}

\newintentry {\hfill $\dss \int\tan u\, du = -\ln |\cos u|$\hfill}

\newintentry {\hfill $\dss \int\cot u\, du= \ln |\sin u|$\hfill}

\newintentry {\hfill $\dss \int\sec u\, du = \ln |\sec u +\tan u|$\hfill}

\newintentry {\hfill $\dss \int\csc u\, du = \ln |\csc u - \cot u|$\hfill}

\newintentry {\hfill $\dss \int {du\over a^2+u^2}={1\over a}\arctan \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {du\over\sqrt {a^2-u^2}}=\arcsin \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int u\, dv=uv-\int v\,du$\hfill}

\vskip 24pt
\centerline {{\bf Integrals involving $au+b$}}
\vskip 18pt

\newintentry {\hfill $\dss \int (au+b)^ndu={(au+b)^{n+1}\over (n+1)a},\quad n\ne -1$\hfill}

\newintentry {\hfill $\dss \int {du\over au+b}={1\over a}\ln |au+b|$\hfill}

\newintentry {\hfill $\dss \int {u\, du\over au+b}={u\over a}-{b\over a^2}\ln |au+b|$\hfill}

\newintentry {\hfill $\dss \int {u\, du\over (au+b)^2}={b\over a^2(au+b)}+{1\over a^2}\ln
|au+b|$\hfill}

\newintentry {\hfill $\dss \int {du\over u(au+b)}={1\over b}\ln \biggl |{u\over au+b}
\biggr |$\hfill}

\newintentry {\hfill $\dss \int u\sqrt {au+b}\,du={2(3au-2b)\over 15a^2}(au+b)^{3/2}$\hfill}

\newintentry {\hfill $\dss \int {u\, du\over\sqrt {au+b}}={2(au-2b)\over 3a^2}\sqrt {au+b}$\hfill}

\newintentry {\hfill $\dss \int u^2\sqrt {au+b}\,du={2\over 105a^3}\left (8b^2-12abu+15a^2u^2\right )
(au+b)^{3/2}$\hfill}

\newintentry {\hfill $\dss \int {u^2\,du\over\sqrt {au+b}}={2\over 15a^3}\left (8b^2-4abu+3a^2
u^2\right )\sqrt {au+b}$\hfill}

\vskip 24pt
\centerline{{\bf Integrals involving $u^2\pm a^2$}}
\vskip 18pt

\newintentry {\hfill $\dss \int {du\over u^2-a^2}={1\over 2a}\ln \biggl |{u-a\over u+a}
\biggr |$\hfill}

\newintentry {\hfill $\dss \int {u\, du\over u^2\pm a^2}=\frac 1/2\ln\left |u^2\pm a^2\right |$\hfill}

\newintentry {\hfill $\dss \int {u^2\, du\over u^2-a^2}=u+{a\over 2}\ln \biggl |{u-a\over
u+a}\biggr |$\hfill}

\newintentry {\hfill $\dss \int {u^2\, du\over u^2+a^2}=u-a\arctan\left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {du\over u(u^2\pm a^2)}=\pm {1\over 2a^2}\ln\biggl |
{u^2\over u^2\pm a^2}\biggr |$\hfill}

\vskip 24pt
\centerline {{\bf Integrals involving $\sqrt {u^2\pm a^2}$}}
\vskip 18pt

\newintentry {\hfill $\dss \int {u\, du\over\sqrt {u^2\pm a^2}}=\sqrt {u^2\pm a^2}$\hfill}

\newintentry {\hfill $\dss \int u\sqrt {u^2\pm a^2}\,du={1\over 3}\left (u^2\pm a^2\right )^{3/2}$\hfill}

\newintentry {\hfill $\dss \int {du\over\sqrt {u^2\pm a^2}}=\ln\Bigl | u+\sqrt {u^2\pm
a^2}\,\Bigr |$\hfill}

\newintentry {\hfill $\dss \int {u^2\,du\over\sqrt {u^2\pm a^2}}={u\over 2}\sqrt {u^2\pm
a^2}\mp {a^2\over 2}\ln\Bigl |u+\sqrt {u^2\pm a^2}\,\Bigr |$\hfill}

\newintentry {\hfill $\dss \int {du\over u\sqrt {u^2+a^2}}={1\over a}\ln \biggl |
{u\over a+\sqrt {u^2+a^2}}\biggr |$\hfill}

\newintentry {\hfill $\dss \int {du\over u\sqrt {u^2-a^2}}={1\over a}\asec \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {du\over u^2\sqrt {u^2\pm a^2}}=\mp {\sqrt {u^2\pm a^2}
\over a^2u}$\hfill}

\newintentry {\hfill $\dss \int\sqrt {u^2\pm a^2}\,du={u\over 2}\sqrt {u^2\pm a^2}\pm
{a^2\over 2}\ln\Bigl |u+\sqrt {u^2\pm a^2}\,\Bigr |$\hfill}

\newintentry {\hfill $\dss \int u^2\sqrt {u^2\pm a^2}\,du={u\over 4}\left (u^2\pm a^2\right )^{3/2}\mp
{a^2u\over 8}\sqrt {u^2\pm a^2}-{a^4\over 8}\ln\Bigl |u+\sqrt {u^2\pm a^2}\,
\Bigr |$\hfill}

\newintentry {\hfill $\dss \int {\sqrt {u^2+a^2}\over u}\,du=\sqrt {u^2+a^2}-a\ln\Biggl |
{a+\sqrt {u^2+a^2}\over u}\,\Biggr |$\hfill}

\newintentry {\hfill $\dss \int {\sqrt {u^2-a^2}\over u}\,du=\sqrt {u^2-a^2}-a\asec \left ({u
\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {\sqrt {u^2\pm a^2}\over u^2}\,du=-{\sqrt {u^2\pm a^2}\over
u}+\ln\Bigl |u+\sqrt {u^2\pm a^2}\,\Bigr |$\hfill}

\vskip 24pt
\centerline {{\bf Integrals involving $\sqrt {a^2-u^2}$}}
\vskip 18pt

\newintentry {\hfill $\dss \int\sqrt {a^2-u^2}\,du={u\over 2}\sqrt {a^2-u^2}+{a^2\over
2}\asin \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {u\, du\over\sqrt {a^2-u^2}}=-\sqrt {a^2-u^2}$\hfill}

\newintentry {\hfill $\dss \int u\sqrt {a^2-u^2}\, du=-{1\over 3}\left (\,a^2-u^2\,\right )^
{3/2}$\hfill}

\newintentry {\hfill $\dss \int {\sqrt {a^2-u^2}\over u}\,du=\sqrt {a^2-u^2}-a\ln
\left |\,{a+\sqrt {a^2-u^2}\over u}\,\right |$\hfill}

\newintentry {\hfill $\dss \int {du\over u\sqrt {a^2-u^2}}=-{1\over a}\ln\left |\,
{a+\sqrt {a^2-u^2}\over u}\,\right |$\hfill}

\newintentry {\hfill $\dss \int u^2\sqrt {a^2-u^2}\,du=-{u\over 4}\left (\,a^2-u^2\,\right 
)^{3/2}+{a^2u\over 8}\sqrt {a^2-u^2}+{a^4\over 8}\asin \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {\sqrt {a^2-u^2}\over u^2}\,du=-{\sqrt {a^2-u^2}\over u}-
\asin \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {u^2\, du\over\sqrt {a^2-u^2}}=-{u\over 2}\sqrt {a^2-u^2}+
{a^2\over 2}\asin \left ({u\over a}\right )$\hfill}

\newintentry {\hfill $\dss \int {du\over u^2\sqrt {a^2-u^2}}=-{\sqrt {a^2-u^2}\over a^2 
u}$\hfill}

 
\vskip 24pt
\centerline {{\bf Integrals involving trigonometric functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int\sin^2 (au)\,du={u\over 2}-{\sin (2au)\over 4a}$\hfill}

\newintentry {\hfill $\dss \int\cos^2 (au)\,du={u\over 2}+{\sin (2au)\over 4a}$\hfill}

\newintentry {\hfill $\dss \int\sin^3 (au)\,du={1\over a}\biggl (\,{\cos^3 (au)\over 3}-
\cos (au)\,\biggr )$\hfill}

\newintentry {\hfill $\dss \int\cos^3 (au)\,du={1\over a}\biggl (\,\sin (au)-{\sin^3 (au)\over
3}\,\biggr )$\hfill}

\newintentry {\hfill $\dss \int\sin^2 (au)\,\cos^2 (au)\,du={u\over 8}-{1\over 32a}\sin 
(4au)$\hfill}

\newintentry {\hfill $\dss \int\tan^2 (au)\,du={1\over a}\tan (au)-u$\hfill}

\newintentry {\hfill $\dss \int\cot^2 (au)\,du=-{1\over a}\cot (au)-u$\hfill}

\newintentry {\hfill $\dss \int\sec^3 (au)\,du={1\over 2a}\sec (au)\tan (au)+{1\over 2a}\ln
|\,\sec (au) +\tan (au)\,|$\hfill}

\newintentry {\hfill $\dss \int\csc^3 (au)\,du=-{1\over 2a}\csc (au)\cot (au)+{1\over 2a}\ln
|\,\csc (au)-\cot (au)\,|$\hfill}

\newintentry {\hfill $\dss \int u\sin (au)\,du={1\over a^2}\left (\sin (au)-au\cos (au)\right )$\hfill}

\newintentry {\hfill $\dss \int u\cos (au)\,du={1\over a^2}\left (\cos (au)+au\sin (au)\right )$\hfill}

\newintentry {\hfill $\dss \int u^2\sin (au)\,du={1\over a^3}\left ( 2au\sin (au)- (\,
a^2u^2-2\,)\cos (au)\right )$\hfill}

\newintentry {\hfill $\dss \int u^2\cos (au)\,du={1\over a^3}\left ( 2au\cos (au)+ (\,
a^2u^2-2\,)\sin (au)\right )$\hfill}

\newintentry {\hfill $\dss \int\sin (au)\,\sin (bu)\,du={\sin (a-b)u\over 2(a-b)}-
{\sin (a+b)u\over 2(a+b)},\qquad a^2\ne b^2$\hfill}

\newintentry {\hfill $\dss \int\cos (au)\,\cos (bu)\,du={\sin (a-b)u\over 2(a-b)}+
{\sin (a+b)u\over 2(a+b)},\qquad a^2\ne b^2$\hfill}

\newintentry {\hfill $\dss \int\sin (au)\,\cos (bu)\,du=-{\cos (a-b)u\over 2(a-b)}-
{\cos (a+b)u\over 2(a+b)},\qquad a^2\ne b^2$\hfill}

\newintentry {\hfill $\dss \int\sin^nu\,du=-{1\over n}\sin^{n-1}u\cos u+{n-1\over n}\int
\sin^{n-2}u\,du$\hfill}

\vskip 24pt
\centerline{{\bf Integrals involving hyperbolic functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int\sinh (au)\,du={1\over a}\cosh (au)$\hfill}

\newintentry {\hfill $\dss \int\sinh^2 (au)\, du={1\over 4a}\sinh (2au)-{u\over 2}$\hfill}

\newintentry {\hfill $\dss \int\cosh (au)\, du={1\over a}\sinh (au)$\hfill}

\newintentry {\hfill $\dss \int\cosh^2 (au)\, du={u\over 2}+{1\over 4a}\sinh (2au)$\hfill}

\newintentry {\hfill $\dss \int\sinh (au)\,\cosh (bu)\, du={\cosh\left ( (a+b)\,u\right )\over 2(a+b)}+
              {\cosh\left ( (a-b)\,u\right )\over 2(a-b)}$\hfill}

\newintentry {\hfill $\dss \int\sinh (au)\,\cosh (au)\, du={1\over 4a}\cosh (2au)$\hfill}

\newintentry {\hfill $\dss \int\tanh u\, du=\ln (\cosh u)$\hfill}

\newintentry {\hfill $\dss \int\sech u\, du=\atan (\sinh u)=2\arctan\left (e^u\right )$\hfill}


\vskip 24pt
\centerline{{\bf Integrals involving exponential functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int ue^{au}\,du={e^{au}\over a^2}(au-1)$\hfill}

\newintentry {\hfill $\dss \int u^2e^{au}\,du={e^{au}\over a^3}\left (\,a^2u^2-2au+2\,\right
)$\hfill}

\newintentry {\hfill $\dss \int u^ne^{au}\,du={1\over a}u^ne^{au}-{n\over a}\int u^{n-1}
e^{au}\,du$\hfill}

\newintentry {\hfill $\dss \int e^{au}\sin (bu)\,du={e^{au}\over a^2+b^2}\left (a\sin (bu)-b\cos 
(bu)\right )$\hfill}

\newintentry {\hfill $\dss \int e^{au}\cos (bu)\,du={e^{au}\over a^2+b^2}\left (a\cos (bu)+b\sin 
(bu)\right )$\hfill}
 
\vskip 24pt
\centerline{{\bf Integrals involving inverse trigonometric functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int\asin\left ({u\over a}\right )\,du=u\asin\left ({u\over a}\right ) +\sqrt {a^2-u^2}$\hfill}

\newintentry {\hfill $\dss \int\acos\left ({u\over a}\right )\,du=u\acos\left ({u\over a}\right ) -\sqrt {a^2-u^2}$\hfill}

\newintentry {\hfill $\dss \int\atan\left ({u\over a}\right )\,du=u\atan\left ({u\over a}\right ) -
{a\over 2}\ln\left (a^2+u^2\right )$\hfill}

\vskip 24pt
\centerline{{\bf Integrals involving inverse hyperbolic functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int\asinh\left ({u\over a}\right )\,du=u\asinh\left ({u\over a}\right )-\sqrt {u^2+a^2}$\hfill}

\newintentry {\hfill \begin{eqnarray*} \dss 
                                    \int\acosh\left ({u\over a}\right )\,du&=
                                    u\acosh\left ({u\over a}\right )-\sqrt {u^2-a^2}\qquad \acosh\left ({u\over a}\right )>0;\\
                      &=  u\acosh\left ({u\over a}\right )+\sqrt {u^2-a^2}\qquad \acosh\left ({u\over a}\right )<0.
\end{eqnarray*}\hfill}

\newintentry {\hfill $\dss \int\atanh\left ({u\over a}\right )\,du=u\atanh\left ({u\over a}\right ) +
{a\over 2}\ln\left (a^2-u^2\right )$\hfill}




\vskip 24pt
\centerline{{\bf Integrals involving logarithm functions}}
\vskip 18pt

\newintentry {\hfill $\dss \int\ln u\,du=u(\ln u-1)$\hfill}

\newintentry {\hfill $\dss \int u^n\ln u\,du=u^{n+1}\biggl\lbrack\,{\ln u\over n+1}-
{1\over (n+1)^2}\,\biggr\rbrack,\qquad n\ne -1$\hfill}

\vskip 24pt    
\centerline{{\bf Wallis' Formulas}}
\vskip 18pt

\newintentry {\hfill \begin{eqnarray*}\dss \int_0^{\pi /2}\sin^mx\,dx&=
\int_0^{\pi /2}\cos^mx\,dx\\
&={(m-1)(m-3)\ldots (2\,\hbox{or}\, 1)\over m(m-2)\ldots (3\,\hbox{or}\, 
2)}k,
\end{eqnarray*}
\hfill}
\vskip 12pt

where $k=1$ if $m$ is odd and $k=\pi /2$ if $m$ is even.
\vskip 18pt

\newintentry {$\dss \int_0^{\pi /2}\sin^mx\,\cos^nx\,dx=$}

\hfill$\dss{(m-1)(m-3)\ldots (2\,\hbox {or}\, 1)(n-1)(n-3)\ldots (2\, \hbox {or}\, 1)
\over (m+n)(m+n-2)\ldots (2\,\hbox{or}\, 1)}\,k,$\hfill
\vskip 12pt

where $k=\pi /2$ if both $m$ and $n$ are even and $k=1$ otherwise. 

